Question

Photoelectric emission is observed from a metallic surface for frequencies $v_{1}$ and $v_{2}$ of the incident light $\left(v_{1}>v_{2}\right)$. If the maximum value of kinetic energy of the photoelectrons emitted in the two cases are in the ration $1: n$ then the threshold frequency of the metallic surface is:

• A. $\frac{v_{1}-v_{2}}{n-1}$
• B. $\frac{ n v_{1}-v_{2}}{ n -1}$
• C. $\frac{ n v_{2}-v_{1}}{n-1}$
• D. $\frac{v_{1}-v_{2}}{n}$

Answer 1

Answer: (B)

Photoelectric equation $hv - hv _{ o }= K _{\max }$
$h\left(v_{1}-v_{0}\right)=K_{1} \ldots$ (1)
(1) and $h \left(v_{2}-v_{0}\right)= K \ldots$(2)
from eq (1)$\&(2)$ we get
$\frac{v_{1}-v_{0}}{v_{2}-v_{0}}=\frac{1}{n}$
$\Rightarrow n v_{1}-n v_{0}=v_{2}-v_{0}$
$nv _{1}-v_{2}= nv _{ o }-v_{ o }$
$\frac{\left( n v_{1}-v_{2}\right)}{( n -1)}=v_{ o }$